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OverLord2011 [107]
2 years ago
11

A model rocket has an acceleration of 12 m/s2. A net force of 18 N is acting on the rocket. What is the mass of the rocket? its

a select answer, the answers are A. 30 kg B. 6.0kg C. 1.5 kg and finally D. 0.67 kg
Physics
1 answer:
larisa86 [58]2 years ago
5 0

Answer:

C.) 1.5 kg

Explanation:

Start with the equation:

F_n_e_t=ma

Plug in what you know, and solve:

18=m(12)\\m=1.5kg

Find matching soluation:

C.) 1.5 kg

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The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This
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To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

g = \frac{GM}{(d-R_{CM})^2}

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

R_{CM} = Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

g = \frac{GM}{(d-R_{CM})^2}

g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}

g = 3.4*10^{-5}m/s^2

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

\omega = \frac{2\pi}{T}

At the same time we have that centripetal acceleration is given as

a_c = \omega^2 r

Replacing

a_c = (\frac{2\pi}{T})^2 r

a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)

a_c = 3.34*10^{-5}m/s^2

3 0
3 years ago
A=i+j-k<br>b=j+k<br><img src="https://tex.z-dn.net/?f=a%20%3D%20i%20%2B%20j%20-%20k%20%20%20%5C%5C%20b%20%3D%20j%20%2B%20k" id="
olga nikolaevna [1]

Answer:

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Explanation:

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3 years ago
Earth's gravitational force just got three times stronger! What happens to your weight?
kicyunya [14]

Your "weight" is the name you give to that gravitational force.
So your question actually says:

           "Your weight just got three times stronger !
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The force F⃗ pulling the string is constant; therefore the magnitude of the angular acceleration α of the wheel is constant for
abruzzese [7]

Answer:

The answer is "\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}"

Explanation:

\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta    - \omega_{0}^{r} \\\\\to \omega^{r} = 2  (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}

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